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Let $S=\{1,2,3,4,5,6\}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties:
$i$. $R$ has exactly $6$ elements.
$ii$. For each $(a, b) \in R$, we have $|a-b| \geq 2$.
Let $Y=\{R \in X$ : The range of $R$ has exactly one element $\}$ and $Z=\{R \in X: R$ is a function from $S$ to $S\}$.
Let $n(A)$ denote the number of elements in a Set $A$.
(There are two questions based on $PARAGRAPH " 1 "$, the question given below is one of them)
($1$) If $n(X)={ }^m C_6$, then the value of $m$ is. . . .
($2$) If the value of $n(Y)+n(Z)$ is $k^2$, then $|k|$ is. . . .
Give the answer or quetion ($1$) and ($2$)
$20,36$
$20,38$
$20,40$
$20,45$
Solution
$\begin{array}{l}|a-b| \geq 2 \text { or }|b-a|=2 \\ \text { Total } \\ a =1 \quad b =3,4,5,6 \quad 8 \\ a =2 \quad b =4,5,6 \quad 6 \\ a =3 \quad b =5,6 \quad 4 \\ a =4 \quad b =6 \quad 2 \\ \overline{\operatorname{sum}=20} \\ n ( X )={ }^{20} C _6={ }^{ m } C _6 \\ m =20 \\\end{array}$
given $|a-b| \geq 2$ so if
$a=1 b=3,4,5,6 \rightarrow 4 \times 2=8$
$a=1 b=4,5,6 \rightarrow 3 \times 2=6$
$a=1 b=5,6 \rightarrow 2 \times 2=4$
$a=1 b=6 \rightarrow 2 \times 1=2$
$i.e. $Total elements in $X$ is ${ }^{20} C _6$
Now for $n ( Y )$,
range of $R$ has exactly one element $i.e$. second elements must be constant in $R$ and since $R$ must have 6 element so it is not possible to satisfy both condition so $n ( Y )=0$.
$n ( z ) \quad 1 \rightarrow 3,4,5,6$
$2 \rightarrow 4,5,6$
$3 \rightarrow 1,5,6$
$4 \rightarrow 1,2,6$
$5 \rightarrow 1,2,3$
$6 \rightarrow 1,2,3,4$
no. of relation that are function will be
$={ }^4 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^4 C _1$
$=(4 \times 3 \times 3)^2= k ^2$
i.e. $k =36$
